Клеймо на всю жизнь

[abbra]

Не хотите больше быть программистом? Оставьте свой след в истории:
http://blog.8thlight.com/dariusz-pasciak/2012/07/03/dereferencing-null-pointer-without-a-seg-fault.html

После того, как прочитали, заходите под кат и ответьте на вопрос: Скажите, а вас учили, что оператор -> над классами в C++ эквивалентен разыменованию указателя в C?

Что интересно, в стандарте С++ в 5.2.5 (Class member access) в третьем абзаце написано:

If E1 has the type “pointer to class X,” then the expression E1->E2 is converted to the equivalent form (*(E1)).E2; the remainder of 5.2.5 will address only the first option (dot).

Правда, 13.5.6 (Class member access) говорит уже более аккуратно об этом:

An expression x->m is interpreted as (x.operator->())->m for a class object x of type T if T::operator->() exists and if the operator is selected as the best match function by the overload resolution mechanism (13.3)

Comments

[ihar hrachyshka (from livejournal.com)]

[C++11: 5.2.5/2]: For the first option (dot) the first expression shall have complete class type. For the second option (arrow) the first expression shall have pointer to complete class type. The expression E1->E2 is converted to the equivalent form (*(E1)).E2; the remainder of 5.2.5 will address only the first option (dot). In either case, the id-expression shall name a member of the class or of one of its base classes. [ Note: because the name of a class is inserted in its class scope (Clause 9), the name of a class is also considered a nested member of that class. —end note ] [ Note: 3.4.5 describes how names are looked up after the . and -> operators. —end note ]

[abbra]

Это написано и в основном посте.